in which b denotes the width of the cross section, and dM denotes the moment created by axial forces applied at the cross-sectional area (b-dy). Integrating this expression over y = -(h/2) to (h/2) and equating it to Mb, we find:

Note that the maximal axial stress is tremendous in comparison with the weight W divided by the cross-sectional area (bh). We could express the weight of the beam in terms of the mass density p times its volume and come up with yet another expression for ao:

Thus, maximum axial stress increases with the square of the length of the beam. It is also inversely proportional to the thickness of the beam.

In reinforced concrete structures, the iron bars that are embedded in concrete resist much of the tension. Concrete can withstand large compressive forces but fractures easily under tension. Thus, when one constructs a concrete beam of the type shown in Fig. 5.16, it must be reinforced at the upper section of the cross section by iron rods. In the human body, when we abduct our arm sideways to the horizontal configuration and keep it there, the tension in the deltoid muscle times its lever arm balances the moment created at the shoulder by the weight of the arm. The internal forces of the human body during pause, movement, and motion are the subject of the next chapter.

Example 5.9. Static Equilibrium of an Elastic String. An elastic string of negligible weight and length L was attached to stationary points A and B (Fig. 5.17). The string was just about tight in the horizontal position, carrying virtually no tension. When a weight W was attached to the midpoint of the string, the two halves of the string made an angle 6 with the horizontal axis. Young's modulus of the string is E. Determine the angle 6 and the tension in the string.

Solution: The condition of force balance in the vertical direction requires that

Both the tension T and the angle 6 are unknowns. Equations of static equilibrium are not sufficient to determine these parameters, and thus this simple system is statically indeterminate. We consider the material properties of the string to come up with an additional equation. Hooke's law dictates that in which E is the Young's modulus of the string, A is its cross-sectional area, and A is the extension of the string AB. The term (A/L) is the axial

Figure 5.17. A cable of negligible weight supporting a weight W.

strain in the string caused by the placing of the weight W. Using trigonometry we can show that

Eliminating T in Eqns. 5.24a and 5.24b, we derive the following equation for 0:

When the stiffness of the string is large, the angle 0 must be small, and we can express tan 0 and sin 0 in powers of 0:

Hence

The tension T can then be found by using Eqn. 5.24a.

Example 5.10. Rods, Weights, and Cords. The rod AB of length 4a is tied to stationary points through two inextensible strings (Fig. 5.18). A weight W is attached to the rod using five elastic strings. These strings all have the same cross-sectional area A and the same Young's modulus E. The weight of the rod and of the strings are negligible. Determine the tension in each of these strings.

Solution: The free-body diagram of point C shown in the figure requires that

in which the tensions T\, T2, and T3 are the tensions in the elastic strings. The angles 0 and $ are determined by the length parameters a and h. In this equation we have three unknowns. Their values cannot be determined uniquely using equations of statics alone. Thus, we focus on the deformation of elastic strings. Let A be the vertical displacement of point C in the direction of the gravitational force W. We assume that the strings are stiff so that A is small in comparison with the lengths of the strings. The extension of each of the elastic strings can then be defined as a function of A:

A2 = [a2 + (h + A)2]0 5 - [a2 + h2]0 5 - A cos 0 A3 = [4a2 + (h + A)2]0 5 - [4a2 + h2]0 5 - A cos $ Thus, the strains in the strings are e1 = Ai/h = A/h €2 = (A/h) cos2 0 e3 = (A/h) cos2 $ We can express the tension in each string in terms of the strain: T1 = E A (A/h) T2 = E A (A/h) cos2 0 T3 = E A (A/h) cos2 $ Substituting these values into Eqn. 5.26, we obtain: T1 = W/(1 + 2 cos3 0 + 2 cos3 $) T2 = cos2 0 W/(1 + 2 cos3 0 + 2 cos3 $) T3 = cos2 $ W/(1 + 2 cos3 0 + 2 cos3 $)

A point force has a magnitude, direction, and a point of application. In actual life, most loads acting on objects are distributed over a surface or a volume. For example, snow load (kg/m2) is a distributed surface load whereas gravity acts throughout the body of an object. The internal stress in a body may depend strongly on the specific distribution of external loads. For example, the soles of shoes are designed to more uniformly distribute ground forces on the base of the foot so as to avoid stress fracture or soft tissue damage.

Point forces lead to different internal stresses than distributed forces of the same magnitude and direction, at least in a small region surrounding the point of application of the force. In the study of motion and equilibrium of a stiff solid body, however, it is sufficient to replace distributed loads with a single force and or moment. Two-force systems are said to be statically equivalent if (a) the resultant force vectors are equal, and (b) the resultant moment of one force system with respect to a point in space is equal to that of the other. This equivalency relationship reflects the fact that when a force is moved up or down along its line of action, its effect on the motion of the body is not altered.

Example 5.11. Distributed Load Acting on a Beam. The force intensity acting on a beam shown in Fig. 5.19 is given by the following equation:

where w is the load per unit length and has the dimensions of N/m, y is a constant, b is the width of the beam, and x is the distance along the e1 direction measured from point O. The loading is over the entire length L of the beam. Determine the equivalent force system.

Solution: The resultant force of the force system given by Eqn. 5.27 is

The moment created by the two-force system must be equal:

in which X denotes the horizontal distance between point O and the point of application of the equivalent horizontal force. Note that X also represents the horizontal distance from point O to the centroid of the area occupied by the distributed load.

The equivalent force system to a distributed load need not be a point force, but it could be a force couple. A couple is composed of two forces that are equal in magnitude but opposite in direction. Because these forces

do not share the same line of action, they exert a moment on the object but no resultant force. The axial stress distribution on the cross section of the cantilever beam of Example 5.8 could be represented by a force couple.

An object is in static equilibrium if (a) the resultant force acting on the object is equal to zero, and (b) the resultant moment acting on the object with respect to a point fixed in space is equal to zero. Thus, the conditions of equilibrium are

Although static analysis is strictly valid for objects in equilibrium, it could provide reasonable approximations for the forces involved in motions with small accelerations.

Problem 5.1. A woman with a tear in the anterior cruciate ligament of her left knee stands putting her weight on crutches and on her right foot as shown in Fig. P.5.1. She weighs 52 kg and has a height of 1.71 m. In this position her body and her crutches make angles of 63° and 80° with

2F = 0 and 2M = 0 The moment M of a force F about point O is defined as

Figure P.5.1. A woman on crutches.

the horizontal plane, respectively. The distance between the point of application of the ground force on her right foot to her center of mass along the axis of her body is 0.89 m. The horizontal distance between the ground force acting on her right foot and the ground force acting on the crutches is 1.1 m. Determine the vertical ground force acting on the crutches. Assume that her body (and her right foot) is positioned symmetrically between the crutches. Answer: 187 N.

Problem 5.2. Determine the forces carried by the truss elements a, b, and c of the truss shown in Fig. P.5.2. All horizontal elements of the truss have equal lengths.

Problem 5.3. Determine the equivalent point force and point couple systems to the distributed loads shown in Fig. P.5.3. Answer: Let e2 be the unit vector in the direction opposite to the gravitational acceleration and let e3 be the unit vector outward normal to the plane of the paper. Then

Problem 5.4. The man shown in Fig. P.5.4 is performing seated machine rows against a resistance of 60 kg. Determine the force his hands exert on the machine at 0 = —20°. Note that in this exercise both arms move back and forth in unison, and therefore, the two positions shown in the figure correspond to different instants of time. The lever is closest to the man's chest at 0 = —20°. Answer: F = 490.5 N.

Problem 5.5. A circular rod of variable radius is held fixed at points B and D (Fig. P.5.5). An axial force P of 7,000 N is exerted on the rod at

50N/cm

50N/cm

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The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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